For k ∈ ℝ , let the solutions of the equation cos sin - 1 x cot tan - 1 cos sin - 1 x = k , 0 < x < 1 2 be α and β , where the inverse trigonometric functions take only principal values. If the solutions of the equation x 2 - b x - 5 = 0 are 1 α 2 + 1 β 2 and α β , then b k 2 is equal to ______.

Step-by-step solution

Given, cos sin - 1 x cot tan - 1 cos sin - 1 x = k Now simplifying cos sin - 1 x = cos cos - 1 1 - x 2 = 1 - x 2 So, cos sin - 1 x cot tan - 1 cos sin - 1 x = k becomes cos sin - 1 x cot tan - 1 1 - x 2 = k And now solving cot tan - 1 1 - x 2 = cotcot - 1 1 1 - x 2 = 1 1 - x 2 So, cos sin - 1 x cot tan - 1 1 - x 2 = k becomes cos sin - 1 x 1 - x 2 = k Now solving cos sin - 1 x 1 - x 2 = 1 - 2 x 2 1 - x 2 So, 1 - 2 x 2 1 - x 2 = k ⇒ 1 - 2 x 2 = k 2 1 - x 2 ⇒ k 2 - 2 x 2 = k 2 - 1 ⇒ x 2 = k 2 - 1 k